hdu 5090 Game with Pearls（最大匹配）

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1914    Accepted Submission(s): 671

Problem Description

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:

1) Tom and Jerry come up together with a number K.

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N.

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls, …, the Nth tube has exact N pearls.

4) If Jerry succeeds, he wins the game, otherwise Tom wins.

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.

 

Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.

 

Output

For each game, output a line containing either “Tom” or “Jerry”.

 

Sample Input

2 
5 1 
1 2 3 4 5 
6 2 
1 2 3 4 5 5 

 

Sample Output

Jerry 
Tom 

 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）  

题意：Jerry 和 Tom 玩一个游戏 ， 给你 n 个盒子 ， a[ i ] 表示開始时 ，
第 i 个盒子中的小球的个数 。

然后 Jerry 能够在每一个盒子里增加 0 或 k的倍数的小球 ，
  操作完后，Jerry 能够又一次排列 盒子的顺序，终于使 第 i 个盒子中有 i 个小球。 若Jerry能
使终于的盒子变成那样，就输出 “Jerry” ，否则 输出 “Tom” 。
大神的解释：

仅仅只是我写的和他的建图的方式不太一样，我是用了n+1到2*n来建图，这里仅仅是想更easy懂所以附上大神解释原理是一样的。

这是大神解释的报告链接：点击打开链接

刚開始仅仅是一个劲的模拟，可是水平太次没有模拟出来。看了别人的思路才知道能够用最大匹配
还是做题太少啊。

#include<stdio.h>
#include<string.h>
#define M 1100
int path[M][M],vis[M],used[M];
int n,k;
int dfs(int x){
	for(int i=n+1;i<=n*2;i++){
		if(!vis[i] && path[x][i]){
			vis[i]=1;
			if(used[i]==-1 || dfs(used[i])){
				used[i]=x;
				return 1;
			}
		}
	}
	return 0;
}
int main(){
	int t,i,j,a;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&k);
		memset(path,0,sizeof(path));
		for(i=1;i<=n;i++){
			scanf("%d",&a);
			for(j=a;j<=n;j+=k){
				path[i][j+n]=1;//把这个点多能加到的点都与这个点相连一条边 
				path[j+n][i]=1;
			}
		}
		int ans=0;
		memset(used,-1,sizeof(used));
		for(i=1;i<=n;i++){
			memset(vis,0,sizeof(vis));
			ans+=dfs(i);
		}
		if(ans==n) printf("Jerry
");
		else printf("Tom
");
	}
	return 0;
}